3.27.27 \(\int \frac {1+x}{(1+3 x+x^2) \sqrt [3]{1-x^3}} \, dx\) [2627]

3.27.27.1 Optimal result

Integrand size = 25, antiderivative size = 231 \[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {5 \sqrt {3} \sqrt [3]{1-x^3}}{2 \sqrt [3]{2} 5^{2/3}-2 \sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}}\right )}{\sqrt [3]{2} 5^{2/3}}+\frac {\log \left (-\sqrt [3]{2} 5^{2/3}+\sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}\right )}{\sqrt [3]{2} 5^{2/3}}-\frac {\log \left (2^{2/3} \sqrt [3]{5}-2\ 2^{2/3} \sqrt [3]{5} x+2^{2/3} \sqrt [3]{5} x^2+\left (\sqrt [3]{2} 5^{2/3}-\sqrt [3]{2} 5^{2/3} x\right ) \sqrt [3]{1-x^3}+5 \left (1-x^3\right )^{2/3}\right )}{2 \sqrt [3]{2} 5^{2/3}} \]

output

1/10*3^(1/2)*arctan(5*3^(1/2)*(-x^3+1)^(1/3)/(2*2^(1/3)*5^(2/3)-2*2^(1/3)* 
5^(2/3)*x+5*(-x^3+1)^(1/3)))*2^(2/3)*5^(1/3)+1/10*ln(-2^(1/3)*5^(2/3)+2^(1 
/3)*5^(2/3)*x+5*(-x^3+1)^(1/3))*2^(2/3)*5^(1/3)-1/20*ln(2^(2/3)*5^(1/3)-2* 
2^(2/3)*5^(1/3)*x+2^(2/3)*5^(1/3)*x^2+(2^(1/3)*5^(2/3)-2^(1/3)*5^(2/3)*x)* 
(-x^3+1)^(1/3)+5*(-x^3+1)^(2/3))*2^(2/3)*5^(1/3)
 

3.27.27.2 Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.86 \[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {5 \sqrt {3} \sqrt [3]{1-x^3}}{2 \sqrt [3]{2} 5^{2/3}-2 \sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}}\right )+2 \log \left (-\sqrt [3]{2} 5^{2/3}+\sqrt [3]{2} 5^{2/3} x+5 \sqrt [3]{1-x^3}\right )-\log \left (2^{2/3} \sqrt [3]{5}-2\ 2^{2/3} \sqrt [3]{5} x+2^{2/3} \sqrt [3]{5} x^2-5^{2/3} (-1+x) \sqrt [3]{2-2 x^3}+5 \left (1-x^3\right )^{2/3}\right )}{2 \sqrt [3]{2} 5^{2/3}} \]

input

Integrate[(1 + x)/((1 + 3*x + x^2)*(1 - x^3)^(1/3)),x]
 

output

(2*Sqrt[3]*ArcTan[(5*Sqrt[3]*(1 - x^3)^(1/3))/(2*2^(1/3)*5^(2/3) - 2*2^(1/ 
3)*5^(2/3)*x + 5*(1 - x^3)^(1/3))] + 2*Log[-(2^(1/3)*5^(2/3)) + 2^(1/3)*5^ 
(2/3)*x + 5*(1 - x^3)^(1/3)] - Log[2^(2/3)*5^(1/3) - 2*2^(2/3)*5^(1/3)*x + 
 2^(2/3)*5^(1/3)*x^2 - 5^(2/3)*(-1 + x)*(2 - 2*x^3)^(1/3) + 5*(1 - x^3)^(2 
/3)])/(2*2^(1/3)*5^(2/3))
 

3.27.27.3 Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.32 (sec) , antiderivative size = 630, normalized size of antiderivative = 2.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {7279, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x+1}{\left (x^2+3 x+1\right ) \sqrt [3]{1-x^3}} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {1-\frac {1}{\sqrt {5}}}{\left (2 x-\sqrt {5}+3\right ) \sqrt [3]{1-x^3}}+\frac {1+\frac {1}{\sqrt {5}}}{\left (2 x+\sqrt {5}+3\right ) \sqrt [3]{1-x^3}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (5-\sqrt {5}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {8 x^3}{\left (3-\sqrt {5}\right )^3}\right )}{5 \left (3-\sqrt {5}\right )^2}-\frac {\left (5+\sqrt {5}\right ) x^2 \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},1,\frac {5}{3},x^3,-\frac {8 x^3}{\left (3+\sqrt {5}\right )^3}\right )}{5 \left (3+\sqrt {5}\right )^2}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2 \left (5-2 \sqrt {5}\right )} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} 5^{2/3}}-\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2 \left (5+2 \sqrt {5}\right )} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} 5^{2/3}}+\frac {\arctan \left (\frac {\frac {2^{2/3} \sqrt [3]{1-x^3}}{\sqrt [3]{5+2 \sqrt {5}}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} 5^{2/3}}+\frac {\arctan \left (\frac {10^{2/3} \sqrt [3]{5+2 \sqrt {5}} \sqrt [3]{1-x^3}+5}{5 \sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3} 5^{2/3}}-\frac {\log \left (x^3-4 \sqrt {5}+9\right )}{6 \sqrt [3]{2} 5^{2/3}}-\frac {\log \left (x^3+4 \sqrt {5}+9\right )}{6 \sqrt [3]{2} 5^{2/3}}-\frac {\log \left (8 x^3+8 \left (9-4 \sqrt {5}\right )\right )}{6 \sqrt [3]{2} 5^{2/3}}-\frac {\log \left (8 x^3+\left (3+\sqrt {5}\right )^3\right )}{6 \sqrt [3]{2} 5^{2/3}}+\frac {\log \left (\sqrt [3]{2 \left (5-2 \sqrt {5}\right )}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2} 5^{2/3}}+\frac {\log \left (\sqrt [3]{2 \left (5+2 \sqrt {5}\right )}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2} 5^{2/3}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2 \left (5-2 \sqrt {5}\right )} x\right )}{2 \sqrt [3]{2} 5^{2/3}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2 \left (5+2 \sqrt {5}\right )} x\right )}{2 \sqrt [3]{2} 5^{2/3}}\)

input

Int[(1 + x)/((1 + 3*x + x^2)*(1 - x^3)^(1/3)),x]
 

output

-1/5*((5 - Sqrt[5])*x^2*AppellF1[2/3, 1/3, 1, 5/3, x^3, (-8*x^3)/(3 - Sqrt 
[5])^3])/(3 - Sqrt[5])^2 - ((5 + Sqrt[5])*x^2*AppellF1[2/3, 1/3, 1, 5/3, x 
^3, (-8*x^3)/(3 + Sqrt[5])^3])/(5*(3 + Sqrt[5])^2) - ArcTan[(1 - (2*(2*(5 
- 2*Sqrt[5]))^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]*5^(2/3)) 
 - ArcTan[(1 - (2*(2*(5 + 2*Sqrt[5]))^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]]/( 
2^(1/3)*Sqrt[3]*5^(2/3)) + ArcTan[(1 + (2^(2/3)*(1 - x^3)^(1/3))/(5 + 2*Sq 
rt[5])^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]*5^(2/3)) + ArcTan[(5 + 10^(2/3)*(5 
 + 2*Sqrt[5])^(1/3)*(1 - x^3)^(1/3))/(5*Sqrt[3])]/(2^(1/3)*Sqrt[3]*5^(2/3) 
) - Log[9 - 4*Sqrt[5] + x^3]/(6*2^(1/3)*5^(2/3)) - Log[9 + 4*Sqrt[5] + x^3 
]/(6*2^(1/3)*5^(2/3)) - Log[8*(9 - 4*Sqrt[5]) + 8*x^3]/(6*2^(1/3)*5^(2/3)) 
 - Log[(3 + Sqrt[5])^3 + 8*x^3]/(6*2^(1/3)*5^(2/3)) + Log[(2*(5 - 2*Sqrt[5 
]))^(1/3) - (1 - x^3)^(1/3)]/(2*2^(1/3)*5^(2/3)) + Log[(2*(5 + 2*Sqrt[5])) 
^(1/3) - (1 - x^3)^(1/3)]/(2*2^(1/3)*5^(2/3)) + Log[-((2*(5 - 2*Sqrt[5]))^ 
(1/3)*x) - (1 - x^3)^(1/3)]/(2*2^(1/3)*5^(2/3)) + Log[-((2*(5 + 2*Sqrt[5]) 
)^(1/3)*x) - (1 - x^3)^(1/3)]/(2*2^(1/3)*5^(2/3))
 

3.27.27.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

3.27.27.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 5.08 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.37

method	result	size
trager	\(\text {Expression too large to display}\)	\(778\)

input

int((1+x)/(x^2+3*x+1)/(-x^3+1)^(1/3),x,method=_RETURNVERBOSE)
 

output

1/10*RootOf(_Z^3-20)*ln((25*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z* 
RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^2-50*RootOf(RootOf(_Z^3-20)^2+10 
*_Z*RootOf(_Z^3-20)+100*_Z^2)^2*RootOf(_Z^3-20)^2*x-15*RootOf(RootOf(_Z^3- 
20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)^3*x+110*(-x^3+1)^(1/ 
3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20 
)*x+5*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2*x-110*(-x^3+1)^(1/3)*RootOf(RootOf( 
_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)-5*(-x^3+1)^(1/3 
)*RootOf(_Z^3-20)^2+130*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100 
*_Z^2)*x^2+39*RootOf(_Z^3-20)*x^2-60*(-x^3+1)^(2/3)-10*RootOf(RootOf(_Z^3- 
20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x-3*RootOf(_Z^3-20)*x+130*RootOf(Roo 
tOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)+39*RootOf(_Z^3-20))/(x^2+3* 
x+1))+RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*ln(-(25*(-x 
^3+1)^(2/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*RootO 
f(_Z^3-20)^2-150*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)^ 
2*RootOf(_Z^3-20)^2*x-5*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100 
*_Z^2)*RootOf(_Z^3-20)^3*x-60*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_ 
Z*RootOf(_Z^3-20)+100*_Z^2)*RootOf(_Z^3-20)*x+5*(-x^3+1)^(1/3)*RootOf(_Z^3 
-20)^2*x+60*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+ 
100*_Z^2)*RootOf(_Z^3-20)-5*(-x^3+1)^(1/3)*RootOf(_Z^3-20)^2-390*RootOf(Ro 
otOf(_Z^3-20)^2+10*_Z*RootOf(_Z^3-20)+100*_Z^2)*x^2-13*RootOf(_Z^3-20)*...
 

3.27.27.5 Fricas [A] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.40 \[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\frac {1}{30} \cdot 50^{\frac {1}{6}} \sqrt {3} \sqrt {2} \arctan \left (\frac {50^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 50^{\frac {2}{3}} \sqrt {2} {\left (3 \, x^{4} + 8 \, x^{3} + 3 \, x^{2} + 8 \, x + 3\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 50^{\frac {1}{3}} \sqrt {2} {\left (41 \, x^{6} - 11 \, x^{5} + 50 \, x^{4} - 35 \, x^{3} + 50 \, x^{2} - 11 \, x + 41\right )} - 20 \, \sqrt {2} {\left (11 \, x^{5} - 15 \, x^{4} + 15 \, x^{3} - 15 \, x^{2} + 15 \, x - 11\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{30 \, {\left (19 \, x^{6} - 69 \, x^{5} + 30 \, x^{4} - 85 \, x^{3} + 30 \, x^{2} - 69 \, x + 19\right )}}\right ) - \frac {1}{300} \cdot 50^{\frac {2}{3}} \log \left (\frac {50^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (3 \, x^{2} - x + 3\right )} + 50^{\frac {1}{3}} {\left (11 \, x^{4} - 4 \, x^{3} + 11 \, x^{2} - 4 \, x + 11\right )} - 20 \, {\left (2 \, x^{3} - x^{2} + x - 2\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{4} + 6 \, x^{3} + 11 \, x^{2} + 6 \, x + 1}\right ) + \frac {1}{150} \cdot 50^{\frac {2}{3}} \log \left (\frac {50^{\frac {2}{3}} {\left (x^{2} + 3 \, x + 1\right )} - 10 \cdot 50^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 50 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} + 3 \, x + 1}\right ) \]

input

integrate((1+x)/(x^2+3*x+1)/(-x^3+1)^(1/3),x, algorithm="fricas")
 

output

1/30*50^(1/6)*sqrt(3)*sqrt(2)*arctan(1/30*50^(1/6)*sqrt(3)*(2*50^(2/3)*sqr 
t(2)*(3*x^4 + 8*x^3 + 3*x^2 + 8*x + 3)*(-x^3 + 1)^(2/3) + 50^(1/3)*sqrt(2) 
*(41*x^6 - 11*x^5 + 50*x^4 - 35*x^3 + 50*x^2 - 11*x + 41) - 20*sqrt(2)*(11 
*x^5 - 15*x^4 + 15*x^3 - 15*x^2 + 15*x - 11)*(-x^3 + 1)^(1/3))/(19*x^6 - 6 
9*x^5 + 30*x^4 - 85*x^3 + 30*x^2 - 69*x + 19)) - 1/300*50^(2/3)*log((50^(2 
/3)*(-x^3 + 1)^(2/3)*(3*x^2 - x + 3) + 50^(1/3)*(11*x^4 - 4*x^3 + 11*x^2 - 
 4*x + 11) - 20*(2*x^3 - x^2 + x - 2)*(-x^3 + 1)^(1/3))/(x^4 + 6*x^3 + 11* 
x^2 + 6*x + 1)) + 1/150*50^(2/3)*log((50^(2/3)*(x^2 + 3*x + 1) - 10*50^(1/ 
3)*(-x^3 + 1)^(1/3)*(x - 1) - 50*(-x^3 + 1)^(2/3))/(x^2 + 3*x + 1))
 

3.27.27.6 Sympy [F]

\[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {x + 1}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 3 x + 1\right )}\, dx \]

input

integrate((1+x)/(x**2+3*x+1)/(-x**3+1)**(1/3),x)
 

output

Integral((x + 1)/((-(x - 1)*(x**2 + x + 1))**(1/3)*(x**2 + 3*x + 1)), x)
 

3.27.27.7 Maxima [F]

\[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x + 1}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} + 3 \, x + 1\right )}} \,d x } \]

input

integrate((1+x)/(x^2+3*x+1)/(-x^3+1)^(1/3),x, algorithm="maxima")
 

output

integrate((x + 1)/((-x^3 + 1)^(1/3)*(x^2 + 3*x + 1)), x)
 

3.27.27.8 Giac [F]

\[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int { \frac {x + 1}{{\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} + 3 \, x + 1\right )}} \,d x } \]

input

integrate((1+x)/(x^2+3*x+1)/(-x^3+1)^(1/3),x, algorithm="giac")
 

output

integrate((x + 1)/((-x^3 + 1)^(1/3)*(x^2 + 3*x + 1)), x)
 

3.27.27.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1+x}{\left (1+3 x+x^2\right ) \sqrt [3]{1-x^3}} \, dx=\int \frac {x+1}{{\left (1-x^3\right )}^{1/3}\,\left (x^2+3\,x+1\right )} \,d x \]

input

int((x + 1)/((1 - x^3)^(1/3)*(3*x + x^2 + 1)),x)
 

output

int((x + 1)/((1 - x^3)^(1/3)*(3*x + x^2 + 1)), x)